1 Introduction

Let \({\mathbb {C}}^{n \times n}\) be the set of all \(n \times n\) matrices whose entries are in \({\mathbb {C}}\). By matrix polynomial, we mean the matrix-valued function of a complex variable of the form

$$\begin{aligned} M(z)=A_mz^m+A_{m-1}z^{m-1}+\cdots +A_1z+A_0, \end{aligned}$$
(1)

where \(A_i\in {\mathbb {C}}^{n\times n}\) for all \(i=0,1,2,\ldots ,m\). If \(A_m\ne 0\), M(z) is called a matrix polynomial of degree m. A number \(\lambda\) is called an eigenvalue of the matrix polynomial M(z), if there exits a nonzero vector \(X\in {\mathbb {C}}^n\), such that \(M(\lambda )X=0\). The vector X is called an Eigenvector of M(z) associated to the eigenvalue \(\lambda\). It should be noted that each finite eigenvalue of M(z) is a root of the characteristic polynomial \(\det (M(z))\). The polynomial eigenvalue problem is to find an eigenvalue \(\lambda\) and a non-zero vector \(X\in {\mathbb {C}}^n\) such that \(M(\lambda )X=0\). For \(m=1\), it is actually the generalized eigenvalue problem

$$\begin{aligned} AX=\lambda BX, \end{aligned}$$

and in addition, if \(B=I\), we have the standard eigenvalue problem

$$\begin{aligned} AX=\lambda X. \end{aligned}$$

Computing eigenvalues of a matrix polynomial is a hard problem. There are iterative methods to compute these eigenvalues (for reference see [5]). Moreover, when computing pseudospectra of matrix polynomials, which provide information about the global sensitivity of the eigenvalues, a particular region of the (possibly extended) complex plane must be identified that contains the eigenvalues of interest, and bounds clearly help to determine such region (for details see [6]). Therefore, it is useful to find the location of these eigenvalues. Note that, if \(A_0\) is singular, then 0 is an eigenvalue of M(z), and if \(A_m\) is singular, then 0 is an eigenvalue of the matrix polynomial \(z^m M(1/z)\). Therefore, to locate the eigenvalues of these matrix polynomials, we always assume that \(A_0\) and \(A_m\) are non-singular.

Notations:

For a matrix \(A\in {\mathbb {C}}^{n\times n}\), the notation \(A\ge 0\) means "A is positive semidefnite", that is for every vector \(X \in {\mathbb {C}}^n\) we have \(X^*AX\ge 0\). By \(A>0\), we mean "A is positive definite", that is \(X^*AX> 0\) for every \(X \in {\mathbb {C}}^n\). Also in this paper for any two matrices \(A,B \in {\mathbb {C}}^{n \times n}\), the notation \(A\ge B\) means \(A-B\ge 0\). Throughout this paper,\(\left\| . \right\|\) denotes a subordinate matrix norm.

In the theory of distribution of zeros of polynomials with complex coefficients, we have the following result due to Cauchy [4, p. 123]

Theorem A

Let \(P(z)=a_0+a_1z+a_2z^2+\cdots +a_mz^m\) be a polynomial of degree m and

$$\begin{aligned} {\mathcal {M}}=\max \left\{ \left| \frac{a_{m-1}}{a_m}\right| ,\left| \frac{a_{m-2}}{a_m}\right| ,\left| \frac{a_{m-3}}{a_m}\right| ,\ldots ,\left| \frac{a_{1}}{a_m}\right| ,\left| \frac{a_{0}}{a_m}\right| \right\} , \end{aligned}$$

then all the zeros of P(z) lie in the circle \(|z|<1+{\mathcal {M}}\).

As an application of Theorem A, Dehmer [3] proved the following (also see [4, Theorem 27.2])

Theorem B

Let \(P(z)=a_0+a_1z+a_2z^2+\cdots +a_mz^m\), \(a_m\ne 0\), \(m\ge 1\) be a complex polynomial of degree m, such that \(|a_m|>|a_i|\) for all \(i=0,1,\ldots ,m-1\), then all the zeros of P(z) lie in the disk \(|z|< 2\).

Trính et al. [2] extended this result to the matrix polynomials and proved the following:

Theorem C

Let \(M(z)=A_0+A_1z+\cdots +A_mz^m\) be a matrix polynomial, whose coefficients \(A_i\in {\mathbb {C}}^{n\times n}\) satisfying the following dominant property

$$\begin{aligned} \Vert A_m\Vert >\Vert A_{i}\Vert \,\, \forall \, i=0,1,2,\ldots , m-1. \end{aligned}$$

Then each eigenvalue \(\lambda\) of M(z) locate in the open disk

$$\begin{aligned} |\lambda |< 1+ \Vert A_m\Vert \Vert A_m^{-1}\Vert . \end{aligned}$$
(2)

In the same paper, they proved the direct extension of Theorem A to matrix polynomial in the form of the following result:

Theorem D

Let \(M(z)=A_0+A_1z+\cdots +A_mz^m\) be a matrix polynomial of degree m, whose coefficients \(A_k\in {\mathbb {C}}^{n\times n}\).

Then each eigenvalue \(\lambda\) of M(z) satisfies

$$\begin{aligned} |\lambda |< 1+{\mathcal {M}}, \end{aligned}$$
(3)

where

$$\begin{aligned} {\mathcal {M}}=\max _{0\le k\le m-1} \Vert A_k\Vert \Vert A_m^{-1}\Vert . \end{aligned}$$

In this paper, we obtain an annulus containing all the eigenvalues of the matrix polynomial M(z) and use it to obtain a refinement of Theorem D. In this direction, we have the following:

Theorem 1.1

Let \(M(z)=A_mz^m+A_{m-1}z^{m-1}+\cdots +A_1z+A_0\) be a matrix polynomial of degree m, where \(A_k\in {\mathbb {C}}^{n\times n}\) and \(A_0, A_m\) are invertible. If \(\alpha _1, \alpha _2, \ldots , \alpha _m\) are m non-zero real or complex numbers, such that \(\sum _{k=1}^{m}|\alpha _k|\le 1\), then each eigenvalue \(\lambda\) of M(z) satisfies \(r_1 \le |\lambda | \le r_2\), where

$$\begin{aligned} r_1=\min _{1\le k\le m} {\left| \frac{\alpha _k}{\Vert A_k\Vert \Vert A_0^{-1}\Vert } \right| }^{1/k} \end{aligned}$$

and

$$\begin{aligned} r_2=\max _{1\le k\le m} {\left| \frac{1}{\alpha _k} \Vert A_{m-k}\Vert \Vert A_m^{-1}\Vert \right| }^{1/k}. \end{aligned}$$

Remark 1

If we take\(n=1\) and let\(A_i=[a_i], i=0,1,\ldots ,m\), we get a result due to Aziz and Qayoom [1] for the zeros of a polynomial with real or complex coefficients.

Now as a refinement of Theorem D, we have the following:

Theorem 1.2

Let

$$\begin{aligned} M(z)&=A_mz^m+A_pz^p+A_{p-1}z^{p-1}+\cdots +A_1z+A_0\\&=A_mz^m+\sum _{k=0}^{p}A_kz^k, \,\,\,\, 0\le p\le m-1 \end{aligned}$$

be a matrix polynomial of degree m, whose coefficients \(A_k\in {\mathbb {C}}^{n\times n}\) \((k=0,1,2,\ldots ,p,m)\) and let

$$\begin{aligned} {\mathcal {M}}=\max _{0\le k\le p} \Vert A_k\Vert \Vert A_m^{-1}\Vert , \end{aligned}$$

then every eigenvalue \(\lambda\) of M(z) satisfies

$$\begin{aligned} |\lambda | \le {\{{(1+{\mathcal {M}})}^{p+1}-1\}}^{\frac{1}{m}}. \end{aligned}$$
(4)

For \(p=m-1\), we have the following:

Corollary 1

Let

$$\begin{aligned} M(z)&=A_mz^m+A_{m-1}z^{m-1}+\cdots +A_1z+A_0 \end{aligned}$$

be a matrix polynomial of degree m, whose coefficients \(A_k\in {\mathbb {C}}^{n\times n}\) \((k=0,1,2,\ldots ,m)\) and let

$$\begin{aligned} {\mathcal {M}}=\max _{0\le k\le {m-1}} \Vert A_k\Vert \Vert A_m^{-1}\Vert , \end{aligned}$$

then every eigenvalue \(\lambda\) of M(z) satisfies

$$\begin{aligned} |\lambda | \le {\{{(1+{\mathcal {M}})}^{m}-1\}}^{\frac{1}{m}}. \end{aligned}$$
(5)

This result is a refinement of Theorem D.

Remark 2

A result of Aziz and Qayoom [1] for the distribution of zeros of polynomials with real or complex coefficients is a special case of Theorem1.2, when we take\(n=1\) and\(A_i=[a_i]\), \({forall}, i=0,1,2,\ldots , m\).

In particular, if \(\Vert A_m\Vert > \Vert A_i\Vert\), \(\forall \, i=0,1,2,\ldots ,p\), then from Theorem 1.2, we get the following:

Corollary 2

Let

$$\begin{aligned} M(z)&=A_mz^m+A_pz^p+A_{p-1}z^{p-1}+\cdots +A_1z+A_0\\&=A_mz^m+\sum _{k=0}^{p}A_kz^k, \,\,\,\, 0\le p\le m-1 \end{aligned}$$

be a matrix polynomial of degree m, whose coefficients \(A_k\in {\mathbb {C}}^{n\times n}\) \((k=0,1,2,\ldots ,p,m)\) and let \(\Vert A_m\Vert > \Vert A_i\Vert\), \(\forall \, i=0,1,2,\ldots ,p\), then each eigenvalue \(\lambda\) of M(z) satisfies

$$\begin{aligned} |\lambda | \le {\{{(1+\Vert A_m\Vert \Vert A_m^{-1}\Vert )}^{p+1}-1\}}^{\frac{1}{m}}. \end{aligned}$$
(6)

If in Corollary 2, we choose \(p=m-1\), we get the following:

Corollary 3

Let

$$\begin{aligned} M(z)&=A_mz^m+A_{m-1}z^{m-1}+\cdots +A_1z+A_0 \end{aligned}$$

be a matrix polynomial of degree m, whose coefficients \(A_k\in {\mathbb {C}}^{n\times n}\) \((k=0,1,2,\ldots ,m)\) and let \(\Vert A_m\Vert > \Vert A_i\Vert\), \(\forall \, i=0,1,2,\ldots ,m-1\), then each eigenvalue \(\lambda\) of M(z) satisfies

$$\begin{aligned} |\lambda | \le {\{{(1+\Vert A_m\Vert \Vert A_m^{-1}\Vert )}^{m}-1\}}^{\frac{1}{m}}. \end{aligned}$$
(7)

This result is a refinement of Theorem C.

2 Lemmas

Before proving the main results, let us first prove the following lemma:

Lemma 1

Let \(M(z)=A_mz^m+A_{p}z^{p}+\cdots +A_1z+A_0\), \(0\le p\le m-1\) be a matrix polynomial of degree m, where \(A_k\in {\mathbb {C}}^{n\times n}\) and \(A_m\) is invertible. If \(\alpha _1, \alpha _2, \ldots , \alpha _{p+1}\) are \(p+1\) non-zero real or complex numbers, such that \(\sum _{k=1}^{p+1}|\alpha _k|\le 1\), then each eigenvalue \(\lambda\) of M(z) lie in the disk

$$\begin{aligned} |\lambda |\le \max _{1\le k\le p+1}{\left| \frac{1}{\alpha _k} \Vert A_{p-k+1}\Vert \Vert A_m^{-1}\Vert \right| }^{\frac{1}{m-p+k-1}}. \end{aligned}$$

Proof

Let

$$\begin{aligned} r=\max _{1\le k\le p+1}{\left| \frac{1}{\alpha _k}\Vert A_{p-k+1}\Vert \Vert A_m^{-1}\Vert \right| }^{\frac{1}{m-p+k-1}}. \end{aligned}$$

This implies

$$\begin{aligned} r^{m-p+k-1}\ge \frac{1}{|\alpha _k|}\Vert A_{p-k+1}\Vert \Vert A_m^{-1}\Vert , \end{aligned}$$

or

$$\begin{aligned} \left| \alpha _k\right| \ge \Vert A_{p-k+1}\Vert \Vert A_m^{-1}\Vert \frac{1}{r^{m-p+k-1}}, \end{aligned}$$

therefore, we have

$$\begin{aligned} \sum _{k=1}^{p+1}\left| \alpha _k\right|&\ge \sum _{k=1}^{p+1}\Vert A_{p-k+1}\Vert \Vert A_m^{-1}\Vert \frac{1}{r^{m-p+k-1}}\\&=\Vert A_0\Vert \Vert A_m^{-1}\Vert \frac{1}{r^{m}}+\cdots + \Vert A_p\Vert \Vert A_m^{-1}\Vert \frac{1}{r^{m-p}}. \end{aligned}$$

This implies

$$\begin{aligned} \sum _{k=1}^{p+1}\left| \alpha _k\right|&\ge \sum _{j=0}^{p}\Vert A_{j}\Vert \Vert A_m^{-1}\Vert \frac{1}{r^{m-j}}. \end{aligned}$$
(8)

Let \(\lambda\) be any eigenvalue of M(z) and X be corresponding unit eigenvector. If possible, suppose that \(|\lambda |> r\), then we have by using (8)

$$\begin{aligned} \Vert M(\lambda )X\Vert&=\Vert (A_m\lambda ^m+A_{p}\lambda ^{p}+\cdots +A_1\lambda +A_0 )X\Vert \\&\ge {|\lambda |}^m\Vert A_m^{-1}\Vert ^{-1}\Big \{1-\sum _{j=0}^{p}\Vert A_{j}\Vert \Vert A_m^{-1}\Vert \frac{1}{{|\lambda |}^{m-j}}\Big \}\\&>{|\lambda |}^m\Vert A_m^{-1}\Vert ^{-1}\Big \{1-\sum _{j=0}^{p}\Vert A_{j}\Vert \Vert A_m^{-1}\Vert \frac{1}{{r}^{m-j}}\Big \}\\&\ge {|\lambda |}^m\Vert A_m^{-1}\Vert ^{-1}\Big \{1-\sum _{k=1}^{p+1}|\alpha _k|\Big \}\\&\ge 0. \end{aligned}$$

Therefore \(\Vert M(\lambda )X\Vert >0\) for \(|\lambda |>r\). Hence our supposition is wrong and thus each eigenvalue \(\lambda\) of M(z) lies in

$$\begin{aligned} |\lambda |\le r = \max _{1\le k\le p+1}{\left| \frac{1}{\alpha _k} \Vert A_{p-k+1}\Vert \Vert A_m^{-1}\Vert \right| }^{m-p+k+1}. \end{aligned}$$

\(\square\)

3 Proofs of theorems

Proof of 1.1

We have \(M(z)=A_0+A_1z+\cdots +A_mz^m\). Let \(\lambda\) be an eigenvalue of M(z) and \(X\in {\mathbb {C}}^n\) be the corresponding eigenvector of M(z). We first show that each eigenvalue \(\lambda\) of M(z) lie in \(|\lambda |\le r_2.\)

We have

$$\begin{aligned} r_2=\max _{1\le k\le m}{\left| \frac{1}{\alpha _k} \Vert A_{m-k}\Vert \Vert A_m^{-1}\Vert \right| }^{\frac{1}{k}}. \end{aligned}$$
(9)

Applying the case when \(p=m-1\), we have from Lemma 1 that each eigenvalue \(\lambda\) of M(z) lies in

$$\begin{aligned} |\lambda |\le r = \max _{1\le k\le m}{\left| \frac{1}{\alpha _k} \Vert A_{m-k}\Vert \Vert A_m^{-1}\Vert \right| }^{k} = r_2. \end{aligned}$$

Hence we conclude that \(|\lambda |\le r_2\).

Now, consider the matrix polynomial

$$\begin{aligned} M^*(z)=z^mM\left( \frac{1}{z}\right) =A_0z^m+A_1z^{m-1}+\cdots +A_m. \end{aligned}$$

Proceeding similarly as above, we have each eigenvalue \(\lambda\) of \(M^*(z)\) satisfies

$$\begin{aligned} |\lambda |&\le \max _{1\le k\le m}\left| \frac{1}{\alpha _k} \Vert A_k\Vert \Vert A_0^{-1}\Vert \right| ^{\frac{1}{k}}\\&= \frac{1}{\displaystyle {\min _{1\le k\le m}{\left| \frac{\alpha _k}{\Vert A_0^{-1}\Vert \Vert A_k\Vert }\right| }^\frac{1}{k}}}\\&=\frac{1}{r_1}. \end{aligned}$$

Replacing z by 1/z and noting that \(M(z)=z^mM^*\left( \frac{1}{z}\right)\), we conclude that if \(\lambda\) is an eigenvalue of M(z), then

$$\begin{aligned} |\lambda |\ge r_1=\min _{1\le k\le m}{\left| \frac{\alpha _k}{\Vert A_0^{-1}\Vert \Vert A_k\Vert }\right| }^\frac{1}{k}. \end{aligned}$$

This completes the proof of theorem.\(\square\)

Proof of 1.2

We have

$$\begin{aligned} {\mathcal {M}}=\max _{1\le k\le p+1} \Vert A_{p-k+1}\Vert \Vert A_m^{-1}\Vert , \end{aligned}$$

therefore

$$\begin{aligned} \Vert A_{p-k+1}\Vert \Vert A_m^{-1}\Vert \le {\mathcal {M}}\,\,\text {for}\,\, k=1,2,3,\ldots ,p+1. \end{aligned}$$
(10)

Now let us choose

$$\begin{aligned} \alpha _k=\left\{ \frac{{(1+{\mathcal {M}})}^m}{{(1+{\mathcal {M}})}^{p+1}-1}\right\} \left\{ \frac{\Vert A_{p-k+1}\Vert \Vert A_m^{-1}\Vert }{{(1+{\mathcal {M}})}^{m-p+k-1}}\right\} \,\,\text {for}\,\, k=1,2,3,\ldots ,m. \end{aligned}$$
(11)

This implies

$$\begin{aligned} \frac{1}{\alpha _k} \Vert A_{p-k+1}\Vert \Vert A_m^{-1}\Vert =\frac{{(1+{\mathcal {M}})}^{m-p+k-1}\left[ {(1+{\mathcal {M}})}^{p+1}-1\right] }{{(1+{\mathcal {M}})}^m}\,\,\text {for}\,\, k=1,2,3,\ldots ,m. \end{aligned}$$
(12)

Since \(A_{p-k+1}=0\) for \(k=p+2,p+3,\ldots ,m\), therefore \(\alpha _k=0,\,\,\text {for}\,\, k=p+2,p+3,\ldots ,m.\) Hence from (11), we have by using (10)

$$\begin{aligned} \sum _{k=1}^{m}|\alpha _k|&=\sum _{k=1}^{p+1}|\alpha _k|\\&=\sum _{k=1}^{p+1}\left| \left\{ \frac{{(1+{\mathcal {M}})}^m}{{(1+{\mathcal {M}})}^{p+1}-1}\right\} \left\{ \frac{\Vert A_{p-k+1}\Vert \Vert A_m^{-1}\Vert }{{(1+{\mathcal {M}})}^{m-p+k-1}}\right\} \right| \\&=\frac{{(1+{\mathcal {M}})}^m}{{(1+{\mathcal {M}})}^{p+1}-1}\sum _{k=1}^{p+1} \Vert A_{p-k+1}\Vert \Vert A_m^{-1}\Vert \frac{1}{{(1+{\mathcal {M}})}^{m-p}{(1+{\mathcal {M}})}^{k-1}}\\&\le \frac{{(1+{\mathcal {M}})}^m}{{(1+{\mathcal {M}})}^{p+1}-1}\sum _{k=1}^{p+1}\frac{{\mathcal {M}}}{{(1+{\mathcal {M}})}^{m-p}}\frac{1}{{(1+{\mathcal {M}})}^{k-1}}\\&=\left( \frac{{(1+{\mathcal {M}})}^m}{{(1+{\mathcal {M}})}^{p+1}-1}\right) \left( \frac{{\mathcal {M}}}{{(1+{\mathcal {M}})}^{m-p}}\right) \sum _{k=1}^{p+1}\frac{1}{{(1+{\mathcal {M}})}^{k-1}}\\&=\left( \frac{{(1+{\mathcal {M}})}^m}{{(1+{\mathcal {M}})}^{p+1}-1}\right) \left( \frac{{\mathcal {M}}}{{(1+{\mathcal {M}})}^{m-p}}\right) \left( \frac{1-\frac{1}{{({\mathcal {M}}+1)}^{p+1}}}{1-\frac{1}{({\mathcal {M}}+1)}}\right) \\&=\left( \frac{{(1+{\mathcal {M}})}^m}{{(1+{\mathcal {M}})}^{p+1}-1}\right) \left( \frac{{\mathcal {M}}}{{(1+{\mathcal {M}})}^{m-p}}\right) \frac{{(1+{\mathcal {M}})}^{p+1}-1}{{(1+{\mathcal {M}})}^p{\mathcal {M}}}\\&=1. \end{aligned}$$

That is

$$\begin{aligned} \sum _{k=1}^{m}|\alpha _k|\le 1. \end{aligned}$$

Now by Lemma 1 and with the help of (12), we have that each eigenvalue \(\lambda\) of M(z) satisfies

$$\begin{aligned} |\lambda |\le r_2&=\max _{1\le k\le p+1}{\left\{ \frac{1}{\alpha _k} \Vert A_{p-k+1}\Vert \Vert A_m^{-1}\Vert \right\} }^{\frac{1}{m-p+k-1}}\\&=\max _{1\le k\le p+1}{\left\{ \frac{{(1+{\mathcal {M}})}^{m-p+k-1}\left[ {(1+{\mathcal {M}})}^{p+1}-1\right] }{{(1+{\mathcal {M}})}^m}\right\} }^{\frac{1}{m-p+k-1}}\\&=\left( 1+{\mathcal {M}}\right) \max _{1\le k\le p+1}{\left\{ \frac{{(1+{\mathcal {M}})}^{p+1}-1}{{(1+{\mathcal {M}})}^m}\right\} }^{\frac{1}{m-p+k-1}}\\&=\left( 1+{\mathcal {M}}\right) {\left\{ \frac{{(1+{\mathcal {M}})}^{p+1}-1}{{(1+{\mathcal {M}})}^m}\right\} }^{\frac{1}{m}}\\&={\left( {(1+{\mathcal {M}})}^{p+1}- 1\right) }^{\frac{1}{m}}. \end{aligned}$$

This completes the proof of theorem.\(\square\)